Average Error: 0.5 → 0.4
Time: 8.0s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
\[\begin{array}{l} t_0 := \left(n \cdot \pi\right) \cdot 2\\ \frac{\sqrt{t_0}}{\sqrt{k} \cdot {t_0}^{\left(\frac{k}{2}\right)}} \end{array} \]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
\begin{array}{l}
t_0 := \left(n \cdot \pi\right) \cdot 2\\
\frac{\sqrt{t_0}}{\sqrt{k} \cdot {t_0}^{\left(\frac{k}{2}\right)}}
\end{array}
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* (* n PI) 2.0)))
   (/ (sqrt t_0) (* (sqrt k) (pow t_0 (/ k 2.0))))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
double code(double k, double n) {
	double t_0 = (n * ((double) M_PI)) * 2.0;
	return sqrt(t_0) / (sqrt(k) * pow(t_0, (k / 2.0)));
}

Error

Bits error versus k

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.5

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
  2. Simplified0.5

    \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
  3. Using strategy rm
  4. Applied div-sub_binary640.5

    \[\leadsto \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{\sqrt{k}} \]
  5. Applied pow-sub_binary640.4

    \[\leadsto \frac{\color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}}}{\sqrt{k}} \]
  6. Applied associate-/l/_binary640.4

    \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}{\sqrt{k} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]
  7. Simplified0.4

    \[\leadsto \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}{\color{blue}{\sqrt{k} \cdot {\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}}} \]
  8. Using strategy rm
  9. Applied *-un-lft-identity_binary640.4

    \[\leadsto \frac{\color{blue}{1 \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}}{\sqrt{k} \cdot {\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}} \]
  10. Applied times-frac_binary640.5

    \[\leadsto \color{blue}{\frac{1}{\sqrt{k}} \cdot \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}}} \]
  11. Simplified0.5

    \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\frac{\sqrt{2 \cdot \left(n \cdot \pi\right)}}{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}}} \]
  12. Using strategy rm
  13. Applied frac-times_binary640.4

    \[\leadsto \color{blue}{\frac{1 \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}}{\sqrt{k} \cdot {\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}}} \]
  14. Simplified0.4

    \[\leadsto \frac{\color{blue}{\sqrt{\left(n \cdot \pi\right) \cdot 2}}}{\sqrt{k} \cdot {\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}} \]
  15. Simplified0.4

    \[\leadsto \frac{\sqrt{\left(n \cdot \pi\right) \cdot 2}}{\color{blue}{\sqrt{k} \cdot {\left(\left(n \cdot \pi\right) \cdot 2\right)}^{\left(\frac{k}{2}\right)}}} \]
  16. Final simplification0.4

    \[\leadsto \frac{\sqrt{\left(n \cdot \pi\right) \cdot 2}}{\sqrt{k} \cdot {\left(\left(n \cdot \pi\right) \cdot 2\right)}^{\left(\frac{k}{2}\right)}} \]

Reproduce

herbie shell --seed 2021198 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))